What is the answer to 3iz^2-2z=-i ?

The answer to 3iz^2-2z=-i is z=-i,z= 1/3 i

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3iz^2-2z=-i

3 i z^{2} - 2 z = - i

z = - i, z = \frac{1}{3} i

Solution steps

3 i z^{2} - 2 z = - i

Substitute

z = x + y i

3 i (x + y i)^{2} - 2 (x + y i) = - i

Expand

3 i (x + y i)^{2} - 2 (x + y i) : (- 6 x y - 2 x) + i (- 2 y + 3 x^{2} - 3 y^{2})

(- 6 x y - 2 x) + i (- 2 y + 3 x^{2} - 3 y^{2}) = - i

Rewrite

- i

in standard complex form:

0 - i

(- 6 x y - 2 x) + i (- 2 y + 3 x^{2} - 3 y^{2}) = 0 - i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[- 6 x y - 2 x = 0 - 2 y + 3 x^{2} - 3 y^{2} = - 1]

[- 6 x y - 2 x = 0 - 2 y + 3 x^{2} - 3 y^{2} = - 1] : (x = 0, x = 0, y = - 1 y = \frac{1}{3})

Substitute back

z = x + y i

z = - i, z = \frac{1}{3} i

x, s i f = y x^{2} - x y^{3}

\frac{( 18 + i )}{4 + 3 i} = a + bi

in (in (in x)) = y

(1 + x) = (1 + r) \cdot (1 + i)

x^{2} - 8 x i - 25 = 0